Ki = 0.5mv^2 = 0.5(0.515)(0.682^2) = 0.11976943J
Kf = 0.5mv^2 = 0.5(0.515)(0.585^2) = 0.0881229375J
73% conservation of kinetic energy.
Little kinetic energy loss to the the lack of a completely frictionless surface and completely elastic collision.
m1 = 1015g = 1.015kg v1 = 0.680m/s
m2 = 551g = 0.551kg v2 = 0.593m/s
m1v1 = m2v2
(1.015kg)(0.680m/s) = (0.551kg)(0.593m/s)
0.690kgm/s = 0.326kgm/s
47% conservation of momentum
Not close enough. I am not entirely sure why this particular video and graph show a significant loss in momentum which is not possible. I would have to guess that the first mass did not collide with the second one completely elastically, thus not transferring all of its momentum. My recollection is that in some of the videos, the blue cup actually recoiled backwards or continued drifting forward for a small distance which would have indicated a lack of complete transfer of momentum. This is also supported by the extreme difference in the final and initial kinetic energy of the system.
Ki = 0.5mv^2 = 0.5(1.015)(0.680^2) = 0.234668J
Kf = 0.5mv^2 = 0.5(0.515)(0.585^2) = 0.0968792995J
41% conservation of kinetic energy
m1 = 515g = 0.515kg v1 = 0.569m/s
m2 = 1030g = 1.03kg v2 = 0.235m/s
(0.515kg)(0.569m/s) = (0.293kgm/s)/(0.515kg + 0.515kg) = 0.285m/s
Real velocity = 0.235m/s
21.276% error
Close enough.
Ki = 0.5mv^2 = 0.5(0.515)(0.569^2) = 0.083368J
Kf = 0.5mv^2 = 0.5(1.03)(0.235^2) = 0.028441J
34% conservation of kinetic energy.
Loss of kinetic energy.
m1 = 515g = 0.515kg v1 = 0.682m/s
m2 = 515g = 0.515kg v2 = 0.585m/s
m1v1 = m2v2
(0.515kg)(0.682m/s) = (0.515kg)(0.585m/s)
0.351kgm/s = 0.301kgm/s
85.76% conservation of momentum.
Close enough.
Ki = 0.5mv^2 = 0.5(0.515)(0.682^2) = 0.11976943J
Kf = 0.5mv^2 = 0.5(0.515)(0.585^2) = 0.0881229375J
73% conservation of kinetic energy.
Little kinetic energy loss to the the lack of a completely frictionless surface and completely elastic collision.
m1 = 1015g = 1.015kg v1 = 0.680m/s
m2 = 551g = 0.551kg v2 = 0.593m/s
m1v1 = m2v2
(1.015kg)(0.680m/s) = (0.551kg)(0.593m/s)
0.690kgm/s = 0.326kgm/s
47% conservation of momentum
Not close enough. I am not entirely sure why this particular video and graph show a significant loss in momentum which is not possible. I would have to guess that the first mass did not collide with the second one completely elastically, thus not transferring all of its momentum. My recollection is that in some of the videos, the blue cup actually recoiled backwards or continued drifting forward for a small distance which would have indicated a lack of complete transfer of momentum. This is also supported by the extreme difference in the final and initial kinetic energy of the system.
Ki = 0.5mv^2 = 0.5(1.015)(0.680^2) = 0.234668J
Kf = 0.5mv^2 = 0.5(0.515)(0.585^2) = 0.0968792995J
41% conservation of kinetic energy
m1 = 515g = 0.515kg v1 = 0.569m/s
m2 = 1030g = 1.03kg v2 = 0.235m/s
(0.515kg)(0.569m/s) = (0.293kgm/s)/(0.515kg + 0.515kg) = 0.285m/s
Real velocity = 0.235m/s
21.276% error
Close enough.
Ki = 0.5mv^2 = 0.5(0.515)(0.569^2) = 0.083368J
Kf = 0.5mv^2 = 0.5(1.03)(0.235^2) = 0.028441J
34% conservation of kinetic energy.
Loss of kinetic energy.
m1 = 1015g = 1.015kg v1 = 0.564m/s
m2 = 1530g = 1.530kg v2 = 0.334m/s
(1.015kg)(0.564m/s) = (0.572kgm/s)/(1.530kg) = 0.374m/s
Real velocity = 0.334m/s
11.976% error
Close enough.
Ki = 0.5mv^2 = 0.5(1.015)(0.564^2) = 0.161434J
Kf = 0.5mv^2 = 0.5(1.530)(0.334^2) = 0.08543J
52.91% conservation of kinetic energy.
Loss of kinetic energy.